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# If x = 7/9 − 15/18 + 10/12 then (1 − x)^2 ?

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Joined: 02 Sep 2009
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If x = 7/9 − 15/18 + 10/12 then (1 − x)^2 ?  [#permalink]

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21 Aug 2018, 06:05
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Difficulty:

5% (low)

Question Stats:

92% (01:20) correct 8% (02:53) wrong based on 48 sessions

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If $$x = \frac{7}{9} − \frac{15}{18} + \frac{10}{12}$$ then $$(1−x)^2$$ ?

A. 1/ 9
B. 4/81
C. 25/144
D. 9/16
E. 25/36

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Re: If x = 7/9 − 15/18 + 10/12 then (1 − x)^2 ?  [#permalink]

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21 Aug 2018, 06:10
Taking LCM of X gives X = 7/9
Then (1-X)^2 = (1-7/9)^2 = 4/81

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If x = 7/9 − 15/18 + 10/12 then (1 − x)^2 ?  [#permalink]

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21 Aug 2018, 11:32
Bunuel wrote:
If $$x = \frac{7}{9} − \frac{15}{18} + \frac{10}{12}$$ then $$(1−x)^2$$ ?

A. 1/ 9
B. 4/81
C. 25/144
D. 9/16
E. 25/36

$$x = \frac{7}{9} − \frac{15}{18} + \frac{10}{12}$$

Or, $$x = \frac{28 - 30 + 30}{36}$$

Or, $$x = \frac{7}{9}$$

So, $$(1−x)^2$$

= $$(1 - \frac{7}{9})^2$$

= $$(\frac{2}{9})^2$$

= $$\frac{4}{81}$$, Answer must be (B)
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Re: If x = 7/9 − 15/18 + 10/12 then (1 − x)^2 ?  [#permalink]

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22 Aug 2018, 04:45
1
Bunuel wrote:
If $$x = \frac{7}{9} − \frac{15}{18} + \frac{10}{12}$$ then $$(1−x)^2$$ ?

A. 1/ 9
B. 4/81
C. 25/144
D. 9/16
E. 25/36

Do not rush into calculations

$$x = \frac{7}{9} − \frac{15}{18} + \frac{10}{12}$$

$$\frac{15}{18}$$=$$\frac{5}{6}$$

$$\frac{10}{12}$$=$$\frac{5}{6}$$

So, $$x = \frac{7}{9} − \frac{5}{6} + \frac{5}{6}$$ = $$\frac{7}{9}$$

$$(1−x)^2$$ = $$(1−\frac{7}{9})^2$$ = $$\frac{4}{81}$$

Re: If x = 7/9 − 15/18 + 10/12 then (1 − x)^2 ?   [#permalink] 22 Aug 2018, 04:45
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